# Math Help - Inequality

1. ## Inequality

Hi !

Either $x,y,z> 0$ , $x + y + z = 1$ . Prove That :

$(1+ \frac{1}{x} ) (1+\frac{1}{y} ) (1+\frac{1}{z} ) \geq 64$

2. Hi !

So, This is My Solution :

$(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z}) \geq 64$
$\Longleftrightarrow$ $(x+1)(y+1)(z+1) \geq 64xyz$
$\Longleftrightarrow$ $xy+yz+zx+2 \geq 63xyz$
by AM-GM :
$xyz \leq \frac{(x+y+z)^3}{27}$
$\Longleftrightarrow$ $54xyz \leq 2$ $(1)$
And We Have : $(x+y+z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) \geq 9$
$\Longleftrightarrow$ $xy+yz+zx \geq 9xyz$ $(2)$
by add $(1)$ and $(2)$ We Have The result sought !

3. Very good.